0=3x^2-62x+119

Solution for 0=3x^2-62x+119 equation:

0=3x^2-62x+119

We move all terms to the left:

0-(3x^2-62x+119)=0

We add all the numbers together, and all the variables

-(3x^2-62x+119)=0

We get rid of parentheses

-3x^2+62x-119=0

a = -3; b = 62; c = -119;
Δ = b2-4ac
Δ = 622-4·(-3)·(-119)
Δ = 2416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2416}=\sqrt{16*151}=\sqrt{16}*\sqrt{151}=4\sqrt{151}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-4\sqrt{151}}{2*-3}=\frac{-62-4\sqrt{151}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+4\sqrt{151}}{2*-3}=\frac{-62+4\sqrt{151}}{-6}
$